Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 18 (a)/Solution 1

Following the hint, let's look for $A,B$ such that $$f(x)=\frac{A}{2+x}+\frac{B}{1-x}.$$

Following steps similar to those of Question 15, we find $A=B=2$. We must now notice that each summand on the right-hand side of $$f(x)=\frac{2}{2+x}+\frac{2}{1-x}$$ is the closed form of a geometric series $\sum _{n=0}^{\mathrm{\infty }}a{x}^n$. Recall that this series converges to $\frac{a}{1-x}$, as long as $x\in (-1,1)$. Let's figure out how to express each summand as a series separately, since we are allowed to recombine them using Theorem 3.5.13 of [CLP].

For the first, we have $$\frac{2}{2+x}=\frac{1}{1+x/2}=\frac{1}{1-(-x/2)}=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{-x}{2}\right)}^n.$$ For the second, we have $$\frac{2}{1-x}=\sum _{n=0}^{\mathrm{\infty }}2x^n,$$ and we will analyse both radii of convergence in part (b) . It follows then that $$f(x)=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{-x}{2}\right)}^n+\sum _{n=0}^{\mathrm{\infty }}2x^n=\sum _{n=0}^{\mathrm{\infty }}(\frac{(-1{)}^n}{2^n}+2)x^n.$$

Therefore the sequence of coefficients is $A_n=(\frac{(-1{)}^n}{2^n}+2)$.